I'm having trouble to solve this problem with the Jacobian method.
Let $\mathbb{I}_{[0,1]^2}(x,y)$ the density of a random vector $(X,Y)$ with $X\perp Y$, i have to find the density of $Z=\frac{X}{Y}$ without the formule $f_{Z}(z)=\int_{-\infty}^{+\infty}|x|f_{X}(x)f_Y(zx)dx$.
After using the transformation $g=\left\{\begin{matrix} u=\frac{x}{y}\\ v=y\end{matrix}\right.$ $\Rightarrow $ $g'=\left\{\begin{matrix} x=uv\\ y=v\end{matrix}\right.$ and calculating the Jacobian matrix (that is $v$ with $dxdy=vdudv$) I have:
$\mathbb{E}[g(\frac{X}{Y})]=\int_{\mathbb{R}^{2}}g(\frac{x}{y})f(x,y)dxdy=\int_{\mathbb{R}^{2}}g(\frac{x}{y})\mathbb{I}_{[0,1]^{2}}(x,y)dxdy=\int_{0}^{1}v[\int_{0}^{1}g(u)du]dv$
But now i don't understand what are the extremes of integration.
Could you help me? Thanks in advance!
The process is purely mechanical.
Pdf of $(X,Y)$ is $$f_{X,Y}(x,y)=\mathbf1_{0<x,y<1}$$
Changing variables $(x,y)\to (u,v)$, you have $$0<x,y<1\implies 0<uv<1\,,\,0<v<1$$
So pdf of $(U,V)$ is just $$f_{U,V}(u,v)=f_{X,Y}(uv,v)|v|=v\mathbf1_{0<uv<1,0<v<1}$$
Hence pdf of $U$ is $$f_U(u)=\int f_{U,V}(u,v)\,dv=\int_0^{\min(1,1/u)}v\,dv=\cdots$$